Question

factorize the polynomial y^3-7y+6​

Answer 1

Answer:

$\large (y-1)(y-2)(y+3)$

Step-by-step explanation:

By hit and trial method .

Put y = 1

$\large p(x)= y^3-7y+6\\\\\\\large p(1)= 1^3-7\times1+6\\\\\\\large p(1)= 7-7\\\\\\\large p(1)=0$

Since remainder comes 0 so ( y - 1 ) is a factor of p ( x ).

On dividing p ( x ) by ( y - 1 ) we get

$\large y^2+y-6$

Now factorise it by splitting mid term

$\large y^2-y-6\\\\\\\large y^2+3y-2y-6\\\\\\\large y(y+3)-2(y+3)\\\\\\\large (y+3)(y-2)$

Thus we factorise the p ( x ) and get factor  $\large (y-1)(y-2)(y+3)$

For zeroes :

y - 1 = 0

y = 1        or   y - 2 = 0

y =  2      or  y + 3 = 0

y = - 3

Answer 2

polynomial, f(x) = x3 + 6x2 + 11x + 6

The constant term in f(x) is 6

The factors of 6 are ± 1, ± 2, ± 3, ± 6

Let, x + 1 = 0

=> x = -1

Substitute the value of x in f(x)

f(-1) = (−1)3 + 6(−1)2 + 11(−1) + 6

= - 1 + 6 - 11 + 6

= 12 – 12

= 0

So, (x + 1) is the factor of f(x)

Similarly, (x + 2) and (x + 3) are also the factors of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

∴ f(x) = k(x + 1)(x + 2)(x + 3)

=> x3 + 6x2 + 11x + 6 = k(x + 1)(x + 2)(x + 3)

Substitute x = 0 on both the sides

=> 0 + 0 + 0 + 6 = k(0 +1)(0 + 2)(0 + 3)

=> 6 = k(1*2*3)

=> 6 = 6k

=> k = 1

Substitute k value in f(x) = k(x + 1)(x + 2)(x + 3)

=> f(x) = (1)(x + 1)(x + 2)(x + 3)

=> f(x) = (x + 1)(x + 2)(x + 3)

∴ x3 + 6x2 + 11x + 6 = (x + 1)(x + 2)(x + 3)